The ratio of the magnetic field at the centre of a current-carrying circular wire and the magnetic field at the centre of a square coil made from the same length of wire will be

$A$ current $I$ enters a circular coil of radius $R$,branches into two parts and then recombines as shown in the circuit diagram. The resultant magnetic field at the centre of the coil is

$A$ magnetic field of $5 \times 10^{-5} \,T$ is produced at a perpendicular distance of $0.2 \,m$ from a long straight wire carrying electric current. If the permeability of free space is $4 \pi \times 10^{-7} \,T \cdot m/A$, the current passing through the wire in $A$ is:

For a wire, as shown in the figure, carrying a current of $10 \,A$, find the magnetic induction field at the point $O$. (Given: $\mu_0 = 4 \pi \times 10^{-7} \,H/m$)

$A$ wire of certain length carries a steady current. It is first bent to form a circular coil of one turn. The same wire is then bent to form a circular coil of three turns. The ratio of magnetic inductions at the centre of the coil in the two cases is

Two circular coils $X$ (smaller) and $Y$ (bigger), each having a single turn, carry equal currents in the same direction and subtend the same angle at point '$O$' along the axis of the coil. The distances of the centers of coil to point '$O$' are $d$ and $(d/2)$ for coil $Y$ and $X$ respectively. The radii of coils $Y$ and $X$ are $(2r)$ and $(r)$ respectively. The magnetic induction due to the bigger coil at point '$O$' is $B_y$ and that due to smaller coil $X$ at point '$O$' is $B_x$. $(d \gg r)$ The relation between $B_x$ and $B_y$ is